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This question is related to the paper http://arxiv.org/abs/1204.5221 and is a continuation of the previous question Symmetries in Wilsonian RG

  • In the liked paper why do the equalities in equation 2.7 and 2.11 hold? (the LHS of both the equations is the same and hence the two equations are 2 different ways of writing the full connected functional W)

    I guess one reads 2.7 to say that when one is flowing down to the IR from UV one develops only "relevant" (dim <4) operators and one I guess reads 2.11 to mean that one develops only irrelevant (dim >4) operators when one flows up to the UV from IR.

    Why?

  • In the linked paper just below equation 2.2 the authors comment that if there is a CFT in the UV then this UV behaviour can change if irrelevant operators are added. why? I would think that (dim>4)/irrelevant operators would come suppressed with positive powers of the cut-off and hence if one pushes the cut-off to infinity then they would vanish and hence the UV is not affected by them. But the authors don't seem to think so...

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1 Answer 1

Regarding your second question: You've got the reasoning backwards (or perhaps a sign wrong in the definitions). Irrelevant deformations of a theory are termed irrelevant because their contributions become less and less important as we zoom out on our field theory. Consequently, they become more and more important as we zoom in. In the UV limit, they are dominant.

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@useer1504 Can you put in your definitions? If O is an irrelevant operator of dimension $4 + n >4$ then would flow in the IR as $O/\Lambda ^n$ where $\Lambda$ is the UV cut-off. Hence in the IR such a high dimension operator will become important since it scales inversely with $\Lambda$. What is wrong in this argument? –  user6818 Apr 11 '13 at 22:48
    
@user6818 $\Lambda$ is a UV momentum cutoff. The IR limit corresponds to $\Lambda \to \infty$. –  user1504 Apr 12 '13 at 13:41
    
Can you explain this some more - as to why I should think the IR limit as taking the UV cut-off to infinity? (..naively I would have thought it to be the reverse!..) –  user6818 Apr 12 '13 at 23:32
    
You're computing the expectation value $\langle \mathcal{O} \rangle$ of an observable which has some characteristic energy scale $\lambda$ (equivalently, a characteristic distance scale $\hbar/\lambda$). The contribution to this observable from an irrelevant operator of dim $4 + n$ will be of order $(\lambda/\Lambda)^n$. Take the IR limit means taking the energy scale of the observable to be very small relative to the energy scale $\Lambda$ of the cutoff. –  user1504 Apr 12 '13 at 23:50
    
@user6818 It occurs to me that the customary notation $\Lambda \to \infty$ is very sloppy because it obscures the critical fact that two distance scales are being compared when one approaches an IR or UV limit. –  user1504 Apr 12 '13 at 23:56

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